Question

Example 5.11 A circular racetrack of
radius 300 m is banked at an angle of 15°
If the coefficient of friction between the
wheels of a race-car and the road is 0.2
what is the (a) optimum speed of the race-
car to avoid wear and tear on its tyres, and
(b) maximum permissible speed to avoid
slipping?​

Answer 1

EXPLANATION.

$\sf : \implies \: a \: circular \: race \: track \: of \: 300m \: \\ \\ \sf : \implies \: banked \: at \: an \: angle \: of \: 15 \degree \\ \\ \sf : \implies \: coefficient \: of \: friction \: between \: the \: wheels \: of \: race \: car \: and \: road = 0.2 \\ \\ \sf : \implies \: \mu \: = 0.2$

$\sf : \implies \: \orange{{ \underline{1) = optimum \: speed \: of \: race \: car \: to \: avoid \: wear \: and \: tear \: on \: its \: tires \: }}} \\ \\ \sf : \implies \: \tan( \theta) = \frac{ {v}^{2} }{rg} \\ \\ \sf : \implies \: {v}^{2} = rg \tan( \theta) \\ \\ \sf : \implies \: {v}^{2} = 300 \times 10 \times \tan(15 \degree) \\ \\ \sf : \implies \: {v}^{2} = 300 \times 10 \times 0.26 \\ \\ \sf : \implies \: {v}^{2} = 780 \\ \\ \sf : \implies \: v \: = \sqrt{780} \approx \: 28$

$\sf : \implies \: \orange{{ \underline{2) = maximum \: permissible \: speed \: to \: avoid \: slipping}}} \\ \\ \sf : \implies \: v_{m} \: = \sqrt{Rg \: ( \frac{ \mu \: + \tan( \theta) }{1 - \mu \: \tan( \theta) } }) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{300 \times 10( \frac{0.2 \times \tan(15 \degree) }{1 - 0.2 \tan( 15 \degree) } } ) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{3000( \frac{0.2 \times 0.26}{1 - 0.2 \times 0.26} } ) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{1458.7} = 38.19 \: ms {}^{ - 1}$

Answer 2

$\huge\bf{\underbrace{\gray{GIVEN:-}}}$

• A circular racetrack of radius 300 m .

• The circular racetrack is banked at an angle of 15° .

• If the coefficient of friction between the wheels of a race-car and the road is 0.2 .

$\huge\bf{\underbrace{\gray{TO\:FIND:-}}}$

1. The optimum speed of the race car to avoid wear and tear on its tyres .
2. The maximum permissible speed to avoid slipping .

$\huge\bf{\underbrace{\gray{SOLUTION:-}}}$

(1) Optimum speed of car is,

$\orange\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{\tan{\theta}\:=\:\dfrac{v^2}{rg}\:}}}}}$

$\bf\pink{:\implies\:v^2\:=\:rg\:\tan{\theta}\:}$

Where,

• $\bf\red{v}$ = Speed of car

• $\bf\red{r}$ = Radius

• $\bf\red{g}$ = gravitational acceleration

• $\bf\red{\theta}$ = Banking of racetrack

☯︎ According to the question,

• $\bf\red{r}$ = 300 m

• $\bf\red{g}$ = 10 m/s²

• $\bf\red{\theta}$ = 15°

$\rm{:\implies\:v^2\:=\:300\times{10}\:\times{\tan{15{\degree}}}\:}$

$\rm{:\implies\:v^2\:=\:780\:}$

$\rm{:\implies\:v\:=\:\sqrt{780}\:}$

$\bf\green{:\implies\:v\:=\:27.9\:m/s\:\:\approx\:28\:m/s\:}$

$\huge\red\therefore$ The optimum speed of the race car to avoid wear and tear on its tyres is "28 m/s" .

__________________________

(2) Maximum permissible speed to avoid slipping is,

$\blue\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{v_m\:=\:\sqrt{rg\:\Big(\dfrac{\mu\:+\:\tan{\theta}}{1\:-\:\mu\:\tan{\theta}}\Big)}\:}}}}}$

Where,

• $\bf\red{\mu}$ = 0.2

$\rm{:\implies\:v_m\:=\:\sqrt{300\times{10}\:\Big(\dfrac{0.2\:+\:\tan{15^{\degree}}}{1\:-\:0.2\times{\tan{15^{\degree}}}}\Big)}\:}$

$\rm{:\implies\:v_m\:=\:\sqrt{1458.7}\:}$

$\bf\green{:\implies\:v_m\:=\:38.19\:m/s\:}$

$\huge\red\therefore$ The maximum permissible speed to avoid slipping is "38.19 m/s" .